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anonymous

  • 4 years ago

PLEASE HELP!!!!!!!! what is log base 3 six square root 243 (where 6 is inside the "v" of the square root"

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  1. anonymous
    • 4 years ago
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    \[\log _{3} \sqrt[6]{243}\]

  2. anonymous
    • 4 years ago
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    \[\log_{3}\sqrt[6]{243} \] like this?

  3. anonymous
    • 4 years ago
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    yes.

  4. anonymous
    • 4 years ago
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    alright lemme see what i can do

  5. anonymous
    • 4 years ago
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    okay, can u help me with a few others as well? i am really not understanding logs :(

  6. anonymous
    • 4 years ago
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    no problem. You know the rule of logs? Where logANS / logBASE = expo?

  7. anonymous
    • 4 years ago
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    ?

  8. anonymous
    • 4 years ago
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    so in this case log [\sqrt[6]{243}\] / log3 = 5/6

  9. anonymous
    • 4 years ago
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    i don't really get it..can you draw it out?

  10. anonymous
    • 4 years ago
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    |dw:1327550139579:dw|

  11. anonymous
    • 4 years ago
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    iam not allowed to use a calculator for these so i have to get it in the exponent form

  12. anonymous
    • 4 years ago
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    \[\frac{1}{6}\log_3(243)\] is a start

  13. anonymous
    • 4 years ago
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    and since \[243=3^5\] you get \[\frac{1}{6}\log_3(3^5)=\frac{5}{6}\]

  14. anonymous
    • 4 years ago
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    |dw:1327550493647:dw|

  15. anonymous
    • 4 years ago
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    was i on the right rack?

  16. anonymous
    • 4 years ago
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    Yes. Now use logs to make the expo turn into a coefficient

  17. anonymous
    • 4 years ago
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    how do i do that ...?

  18. anonymous
    • 4 years ago
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    log3^y = log243^(1/6)

  19. anonymous
    • 4 years ago
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    Now you can put the y in front of the log as well as the 1/6 so it becomes ylog3 = (1/6)[log243]

  20. anonymous
    • 4 years ago
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    i am so confused :'''(

  21. anonymous
    • 4 years ago
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    shall we go off satelite73's method? We keep it as a log form.

  22. anonymous
    • 4 years ago
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    i have to do this non calc

  23. anonymous
    • 4 years ago
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    So from the \[\sqrt[6]{}\] we can turn it into an exponent as 1/6 so\[\log_{3}243^{1/6}\]

  24. anonymous
    • 4 years ago
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    i get that now, so im using satelites method as that does not require a calc.

  25. anonymous
    • 4 years ago
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    really your life would be a lot easier if you put the 1/6 out front

  26. anonymous
    • 4 years ago
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    i said that too

  27. anonymous
    • 4 years ago
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    \[\log(x^n)=n\log(x)\] so \[\log(\sqrt[6]{x})=\frac{1}{6}\log(x)\]

  28. anonymous
    • 4 years ago
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    i am really not understanding anything. i just can't do this :(

  29. anonymous
    • 4 years ago
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    The reason we use logrithims for unknown exponential equations is to turn it from an exponent to a coefficient. It then becomes an algebraic equation. So in this situation we have \[\log_{3}\sqrt[6]{243} \] We turn the \[\sqrt[6]{243}\] into \[243^{1/6}\] So right now we have \[\log_{3} 243^{1/6}\] now since we have the log, we are allowed to drop the 1/6 down from the exponent level to become the leading coefficient thus making the equation \[1/6\log_{3}243 \] From here we can change the number 243 into a number with a base of 3. We know that 3^5 = 243 therefore we can write \[1/6\log_{3}(3^5) \] And knowing that when a base and the number above it are the same number u can cancel out the log and it's base, leaving you with just the 5. So you are to be left with 1/6 * 5 this is known to be 5/6

  30. anonymous
    • 4 years ago
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    dang. thank you so much! alb to help me with anymore?

  31. anonymous
    • 4 years ago
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    Yessir! glad to help :)

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